Question

Given, $y = f(x)$; $\dfrac{d^2 y}{dx^2} + 4y = 0$; $y(0) = 0$; $\dfrac{dy}{dx}(0) = 1$. The problem is a/an

• A. initial value problem having solution $y = x$
• B. boundary value problem having solution $y = x$
• C. initial value problem having solution $y = \dfrac{1}{2}\sin 2x$
• D. boundary value problem having solution $y = \dfrac{1}{2}\sin 2x$

Hint:

If all conditions are given at a single point, it is an initial value problem; if at different points, it becomes a boundary value problem.

Answer 1

The Correct Option is A

The differential equation is
\[ \dfrac{d^2 y}{dx^2} + 4y = 0. \]
Its characteristic equation is
\[ m^2 + 4 = 0 \Rightarrow m = \pm 2i. \]
So the general solution is
\[ y(x) = C_1 \cos 2x + C_2 \sin 2x. \]
Apply the initial conditions:
\[ y(0) = 0 \Rightarrow C_1 = 0. \]
Now,
\[ y'(x) = -2C_1 \sin 2x + 2C_2 \cos 2x = 2C_2 \cos 2x. \]
Given
\[ y'(0) = 1 \Rightarrow 2C_2 = 1 \Rightarrow C_2 = \frac{1}{2}. \]
Thus the solution is
\[ y(x) = \frac{1}{2} \sin 2x. \]
The conditions $y(0)=0$ and $y'(0)=1$ are both specified at the same point, so this is an initial value problem.
Final Answer: (C)